6
$\begingroup$

I currently write an article where I want to introduce roots. Thus I need to motivate them. Here I said, they can be used to find solutions of equations like $x^n=a$. Now I want to make some examples, where this problem arises. So far I found:

  • What is the side length of a square with the area of $10 m^2$?
  • What are the roots of the polynomial $x^2+x-1$?

What examples would you take? I do not have the feeling that mine are really convincing to study roots...

$\endgroup$
  • 4
    $\begingroup$ What is the hypotenuse of the isosceles right triangle with leg length one inch? (Lots of history with this.) $\endgroup$ – Sue VanHattum May 28 '15 at 21:43
  • 2
    $\begingroup$ Are you including e.g. all 0s of polynomials? If so, what about Calculus problems where one wants the derivative (or second derivative, etc) of a polynomial (which is also a polynomial...) to be 0? I do not think there is a shortage of such problems. Maybe you could make your question a bit more specific? $\endgroup$ – Benjamin Dickman May 28 '15 at 22:26
  • $\begingroup$ How about solving for the percentage annual return which will double an investment in ten years? $\endgroup$ – Austin Jun 5 '15 at 4:54
9
$\begingroup$

Ray-tracing, which underlies much of high-end computer graphics (from Toy Story to Frozen) relies on computing the intersection of a ray line-of-sight with a geometric object. For example, where does the ray $a + t v$ with parameter $t$, $a=(2,0)$, $v=(-1,\frac{1}{2})$, intersect the unit circle $x^2 + y^2 = 1$? Substitution of $a + t v = \left( 2-t,\frac{t}{2} \right)$ for $(x,y)$ in the circle equation leads to \begin{eqnarray} (2-t)^2 + \frac{t^2}{4} &=& 1 \;, \\ \frac{5 t^2}{4}-4 t+3 &=& 0 \end{eqnarray} whose roots are $t=2$, or $t=\frac{6}{5}$, yielding the intersection points \begin{eqnarray} (x,y) &=& (0,1) \;, \\ (x,y) &=& \left( \frac{4}{5},\frac{3}{5} \right) &=& (0.8, 0.6) \;. \end{eqnarray}


RayCirc
Because the latter point is closer to the ray origin $a$, that $(0.8, 0.6)$ point is the one whose surface characteristics determine the color/shade of the image pixel represented by the ray line-of-sight.

$\endgroup$
2
$\begingroup$

To compute the golden ratio, we need to find the positive root of a quadratic equation.

Two quantities $a$ and $b$ with $a>b>0$ are said to be in the golden ratio if $a$ is to $b$ as $a+b$ is to $a$.

enter image description here (Image source: Wikipedia)

Algebraically, $$\frac a b= \frac {a+b} a.$$ If $\varphi=\tfrac a b$ denotes the golden ratio, then $$\varphi = 1+\varphi^{-1}.$$ Multiplying with $\varphi$ and rearranging gives the equation $$\varphi^2-\varphi-1=0,$$ which has a positive root $$\varphi=\frac {1+\sqrt 5} 2 \approx 1.618$$

$\endgroup$
  • $\begingroup$ Let $x = -\varphi$. Then $\varphi^2 - \varphi - 1 = x^2 + x - 1$, which amounts to the second bullet-point from the OP. $\endgroup$ – Benjamin Dickman Jun 7 '15 at 5:12
  • $\begingroup$ @BenjaminDickman ... or let $x=b/a=\varphi^{-1}$. Then you also end up with the equation $x^2+x-1=0$. $\endgroup$ – Dag Oskar Madsen Jun 7 '15 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.